What is the integral from 0 to pi/(12) of sin^5(6x) ?

The integral from 0 to pi/(12) of sin^5(6x) is 4/45

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integral from 0 to pi/(12) of sin^5(6x)

\int_{0}^{\frac{π}{12}} sin^{5} (6 x) d x

\frac{4}{45}

Decimal

0.08888 \dots

Solution steps

\int_{0}^{\frac{π}{12}} sin^{5} (6 x) d x

Apply u-substitution

: \int_{0}^{\frac{π}{2}} sin^{5} (u) \frac{1}{6} d u

= \int_{0}^{\frac{π}{2}} sin^{5} (u) \frac{1}{6} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{6} \cdot \int_{0}^{\frac{π}{2}} sin^{5} (u) d u

Rewrite using trig identities

= \frac{1}{6} \cdot \int_{0}^{\frac{π}{2}} (1 - cos^{2} (u))^{2} sin (u) d u

Apply u-substitution

: \int_{1}^{0} - 1 + 2 v^{2} - v^{4} d v

= \frac{1}{6} \cdot \int_{1}^{0} - 1 + 2 v^{2} - v^{4} d v

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= \frac{1}{6} (- \int_{0}^{1} - 1 + 2 v^{2} - v^{4} d v)

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{6} (- (- \int_{0}^{1} 1 d v + \int_{0}^{1} 2 v^{2} d v - \int_{0}^{1} v^{4} d v))

\int_{0}^{1} 1 d v = 1

\int_{0}^{1} 2 v^{2} d v = \frac{2}{3}

\int_{0}^{1} v^{4} d v = \frac{1}{5}

= \frac{1}{6} (- (- 1 + \frac{2}{3} - \frac{1}{5}))

Simplify

\frac{1}{6} (- (- 1 + \frac{2}{3} - \frac{1}{5})) : \frac{4}{45}

= \frac{4}{45}

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What is the integral from 0 to pi/(12) of sin^5(6x) ?
The integral from 0 to pi/(12) of sin^5(6x) is 4/45